How to get the desired result directly by derivation

D[Integrate[t*f[x^2 - t^2], {t, 0, x}], x] == Integrate[D[t*f[x^2 - t^2], x], {t, 0, x}]

is solvable with Mathematica since Integrate can deal with formal integrals if the integral in infinite.

As the short passage about formal integrals in the documentation page shows, prefer integrals without bounds.

Integrate[2 t x Derivative[1][f][-t^2 + x^2], t]

(-x f[-t^2 + x^2])

By hand this can be set into bounds t==0 to t==x:

-xf[0]+xf[x^2]

This problem does not depend on the symbol for the upper bound. This is done with Mathematica V12.0.0.

There is the condition necessary that f[0]==0 to get Your result.

To change the sequence order between integration and differentiation there is another condition necessary: f has to be sufficiently steady and the derivate has to exist on as much points as the function is steady on.

Reasons for that are given in the section Possible Issues for Indefinite Integrals and Definite Integrals. Example:

When part of a sum cannot be integrated explicitly, the whole sum will stay unintegrated:

Integrate[f[x] + f'[x], x]

([Integral](f[x] + Derivative[1][f][x]) [DifferentialD]x)

If entered different:

Integrate[f[x], x] + Integrate[f'[x], x]

(f[x] + [Integral]f[x] [DifferentialD]x)

There is a Mathematica built-in called Derivative that can do derivative with negative integer order:

Derivative[-1][Function[t, 2 x t f'[-t^2 + x^2]]]

(Function[x, -x f[-t^2 + x^2]])

Both together stays unevaluated in the abstract f situation.